0.6
Repeatedly divide by 5 (noting the remainders) until the quotient is zero. Then write the remainders out in reverse order.
3
add all the numbers up and count how many numbers there are then divide it2+3+5+7=17 count the numbers. 4 17 divide by 4 = 4.25
3, 5, 11
2 , 3, 5 ,7
5
10.
Repeatedly divide by 5 (noting the remainders) until the quotient is zero. Then write the remainders out in reverse order.
Say for example you have 16/5.In remainders, the answer would be 3r1. Now you have to divide the remainder by the divisor So for example: 1/5=0.2 Now you have to place the 3 from the answer we got from the remainder in front of the 0.2 answer: 3.2
prime numbers that divide into 1155 = 3, 5, 7, 11
105 can be divided evenly by 1, 3, 5, 7, 15, 21, 35, and 105. These are all the factors of 105, meaning they divide into 105 without leaving a remainder. Additionally, 105 can be divided by other numbers that are not factors, but those divisions will result in remainders.
When 9 is used as a divisor, the remainders can range from 0 to 8. This is because the remainder is always less than the divisor. So, if you divide any number by 9, the possible remainders can be 0, 1, 2, 3, 4, 5, 6, 7, or 8.
One possibility is remainders after division by 5.
The Least Common Denominator (LCD) of 5 and 3 is 15. To find the LCD, you need to determine the smallest number that both 5 and 3 can divide into evenly. In this case, 15 is the smallest number that both 5 and 3 can divide into without any remainders.
All non-negative numbers smaller than 9 ie 0, 1, 2, 3, 4, 5, 6, 7 and 8.
20 / 4 = 5 (no remainder) 21 / 4 = 5 remainder 1 22 / 4 = 5 remainder 2 23 / 4 = 5 remainder 3 24 / 4 = 6 (no remainder) Therefore the numbers that can remain after you've divided a number by four are 0, 1, 2, and 3.
divide both numbers by their common factor (5). 15 divide by 5 is 3 25 divide by 5 is 5 So...... 3 over 5